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Oksi-84 [34.3K]
4 years ago
5

Sandy takes a day off once every 4 days, and Morgan takes a day off once every 10 days. Today, both of them have taken the day o

ff. How many days will it be until they have their next day off together?
Mathematics
2 answers:
Alenkasestr [34]4 years ago
4 0
Lets say that the day they both take off is day 0.
The next day Sandy takes off will be day 4, and Morgans will be day 10.
Sandy will take off 4, 8, 12, 16, and 20.
Morgan will take off 10, 20, 30, 40, and 50.

Both Morgan and Sandy take off day 20, so the next time they have a day off together will be in 20 days.
Lubov Fominskaja [6]4 years ago
4 0
The interval for each time they take the day off would be every 20 days.
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The numerical value of sin²5° + sin²10° + sin²15° +... sin²85° + sin²90° is equal a) 17/2 b) 19/2 c) 15/2 d) 13/2​
Vikki [24]

Answer: b)~\Large\boxed{\frac{19}{2} }

Step-by-step explanation:

<h3>Given expression</h3>

sin²5° + sin²10° + sin²15° +... sin²85° + sin²90°

<h3>Concept:</h3>

sin²x + cos²x = 1

sin(x) = cos (90 - x)

<u />

<u>There are in total these terms:</u>

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90

<u>In total, there are 18 terms, and the first one matches with the second to the last one:</u>

5 -- 85

10 -- 80

.

.

.

40 -- 50

<u />

<u>There are 2 terms left over:</u>

sin²45 and sin²90

<h3>Convert the first half of the sine terms (sin²5 - sin²40) to the cosine terms</h3>

sin²5 = cos² (90 - 5) = cos²85

sin²10 = cos² (90 - 10) = cos²80

.

.

.

sin²40 = cos² (90 - 40) = cos²50

<h3>Simplify the 16 grouped terms </h3>

<em>i.e. sin²85 and cos²85</em>

Using the concept of sin²x + cos²x = 1

sin²85 + cos²85 = 1

sin²80 + cos²80 = 1

.

.

.

sin²50 + cos²50 = 1

Total = (16/2) × 1 = 8 × 1 = 8

<h3>Evaluate the 2 terms that are left over</h3>

sin²45 = (sin45) (sin45) = (√2 / 2) (√2 / 2) = 1/2

sin²90 = (sin90) (sin 90) = (1) (1) = 1

<h3>Add all the terms together</h3>

8+\dfrac{1}{2} +1=\Large\boxed{\frac{19}{2} }

Hope this helps!! :)

Please let me know if you have any questions

8 0
1 year ago
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